The temperature of a certain cup of coffee 10 minutes after it was poured was 120 degrees Fahrenheit. If the temperature F of the coffee t minutes after it was poured can be determined by the formula F = 120(2^(-at)) + 60, where F is in degrees Fahrenheit and a is a constant, then the temperature of the coffee 30mins after it was poured was how many degrees Fahrenheit?
a. 65 b. 75 c. 80 d. 85 e. 90
The temperature of a certain cup of coffee 10 minutes after it was poured was 120 degrees Fahrenheit. If the temperature F of the coffee t minutes after it was poured can be determined by the formula F = 120(2^(-at)) + 60, where F is in degrees Fahrenheit and a is a constant, then the temperature of the coffee 30mins after it was poured was how many degrees Fahrenheit?
a. 65 b. 75 c. 80 d. 85 e. 90
b) 75
Solving for the given condition to get a
120 = 120(2^(-a*10)) + 60
60 = 120(2^(-a*10))
60/120 = (2^(-a*10))
1/2 = 1/ (2^(a*10))
=> a = 1/10
Then subsititute that in the equation for the new time.
F = 120(2^(-at)) + 60
F = 120(2^(- (1/10) * 30)) + 60
F = 120 * 2 ^ -8 + 60
F = 15 + 60
F = 75
Came across this question..in one of the GMAT paper tests...
I don't really find this solvable.. can someone please help?
In the xy-plane, the point (-2, -3) is the center of a circle. The point (-2, 1) lies inside the circle and the point (4, -3) lies outside the circle. If the radius r of the circle is an integer, then r = ?
A. 6
B. 5
C. 4
D. 3
E. 2
The center is (-2,-3) point (-2,1) is inside so r > distance between (-2,-3) and (-2,1) which comes out to be 4. and point (4,-3) is outside so r
Hence r4 so the correct answer is r = 5
Can any one please send me the link for solutions of PS of OG 10?? Thanks
How to find Nth Prime No. I know prime numbers can be written as 6n+1 and 6n-1 apart from 2 and 3 but this way i can't find out the Nth prime no.
Like 5th prime number is 11 but it can be written as 6*2-1 = 11 where n = 2
but here n is not the nth term or prime.
So how to find Nth prime no.
How to find Nth Prime No. I know prime numbers can be written as 6n+1 and 6n-1 apart from 2 and 3 but this way i can't find out the Nth prime no.
Like 5th prime number is 11 but it can be written as 6*2-1 = 11 where n = 2
but here n is not the nth term or prime.
So how to find Nth prime no.
AFAIK, there is no generic formula for derivation of prime numbers.
Originally Posted by deepa7685 View Post
For every positive even integer n, the function h(n) is defined to be the product of all even integers from 2 to n inclusive, if p is the smallest prime factor of h(100) +1, then p is
a) between 2 & 10
b) between 10 & 20
c) between 20 & 30
d) between 30 & 40
e) greater than 40.
Explanations will be appreciated.....
Simple answer would be E
Explanation : Since h(100)+1 can be written as 6n+1 so obviously its a prime no. & the smallest factor other than 1 is the number itself.
So definitely greater than 40.
For every positive even integer n, the function h(n) is defined to be the product of all even integers from 2 to n inclusive, if p is the smallest prime factor of h(100) +1, then p is
a) between 2 & 10
b) between 10 & 20
c) between 20 & 30
d) between 30 & 40
e) greater than 40.
Explanations will be appreciated.....
Lots of explanations for this one. I'll put it in a simple manner.
h(100) = 2.4.6.8.....100 = 2^50(1.2.3.4....50) ---> From the product, you can see that h(100) is divisible by every prime less than 50
So, h(100) + 1 would leave remainder 1 when divided by all of these primes.
Thus, the least prime that divides h(100) + 1 would be greater than 50.
Lots of explanations for this one. I'll put it in a simple manner.
h(100) = 2.4.6.8.....100 = 2^50(1.2.3.4....50) ---> From the product, you can see that h(100) is divisible by every prime less than 50
So, h(100) + 1 would leave remainder 1 when divided by all of these primes.
Thus, the least prime that divides h(100) + 1 would be greater than 50.
I think we need (h(100)+1) 's prime factor.
Please provide solution for the attached question
27943
RagnRoll SaysI think we need (h(100)+1) 's prime factor.
Yes, thats what i have found out. Kindly look at the soln again.
The correct answer is 1/2
You would get the least area when the triangle is an equilateral one.
Side of triangle = 1
Area = _/3/4 * 1^2 = _/3 / 4
So option 1 is correct.
Please provide solution for the attached question
27943
nirjharv SaysThe correct answer is 1/2
Oops, my mistake. I found out the least area, while we had to find the maximum area.
In that case, we take a right angled triangle, centered at the center of the circle.
Area = 1/2 * 1 * 1= 1/2.
This is coz Area of triangle = 1/2 ab Sin(x). where x is the angle between 2 sides of triangle, a and b.
a = b = radius = 1
Sin(x) is maximum for x = 90 degrees. Hence we take a right angled triangle.
The solution is as follows:
area of the required triangle would be 1/2(bcsinx)
Now, sin x is maximum when x=90. hence, max area would be 1/2 since b and c are 1 each
nirjharv SaysThe correct answer is 1/2
You would get the least area when the triangle is an equilateral one.
Side of triangle = 1
Area = _/3/4 * 1^2 = _/3 / 4
So option 1 is correct.
Please provide solution for the attached question
27943
Ditto bro....
... thanks anyways
The solution is as follows:
area of the required triangle would be 1/2(bcsinx)
Now, sin x is maximum when x=90. hence, max area would be 1/2 since b and c are 1 each
Oops, my mistake. I found out the least area, while we had to find the maximum area.
In that case, we take a right angled triangle, centered at the center of the circle.
Area = 1/2 * 1 * 1= 1/2.
This is coz Area of triangle = 1/2 ab Sin(x). where x is the angle between 2 sides of triangle, a and b.
a = b = radius = 1
Sin(x) is maximum for x = 90 degrees. Hence we take a right angled triangle.
nirjharv Saysthe correct answer is 1/2
i think answer should be rt3/4 only.
In all triangles,the equilateral triangle has the greatest possible area.
And then the radius = side of triangle = 1cm.
Correct me if am wrong.
Equilateral triangle will give me the minimum area..i think...
Dude see the solution below:
i think answer should be rt3/4 only.
In all triangles,the equilateral triangle has the greatest possible area.
And then the radius = side of triangle = 1cm.
Correct me if am wrong.
The solution is as follows:
area of the required triangle would be 1/2(bcsinx)
Now, sin x is maximum when x=90. hence, max area would be 1/2 since b and c are 1 each
Equilateral triangle will give me the minimum area..i think...
Dude see the solution below:
BUDDY I HAVE SEEN YOUR SOLUTION,BUT MY DOUBT IS WHAT I MENTIONED IN previous POST OF MINE.
CAN ANYONE CLEAR MY doubt.
@nirjharv- anyways thanks for your effort.:)
i think answer should be rt3/4 only.
In all triangles,the equilateral triangle has the greatest possible area.
And then the radius = side of triangle = 1cm.
Correct me if am wrong.
Well not necessarily, since over here the length of 2 sides is determined to be equal to radius, the determining factor for the area is the angle between those two sides.
So area = AB Sin(x), where x = angle between AB. And A = B = radius = 1.
Sin(x) is max at 90.
If you are looking for maximum area of a triangle inside a circle, such that all vertices are on the circle, then the triangle would be an equilateral one.
@nirjharv: my equilateral soln was a blunder. It doesnt give the min area either. Min area could be a lot lesser.
Well not necessarily, since over here the length of 2 sides is determined to be equal to radius, the determining factor for the area is the angle between those two sides.
.
OK THANKS ANGAD FOR CLEARING MY DOUBT.
ONE MORE QUESTION FROM MY SIDE:
A merchant buys two articles for Rs.600. He sells one of them at a profit of 22% and the other at a loss of 8% and makes no profit or loss in the end. What is the selling price of the article that he sold at a loss?
- Rs. 404.80
- Rs. 440
- Rs. 536.80
- Rs. 160